To prove: (a² – b²), (b² – c²), (c² – d²) are in GP.

Given: a, b, c are in GP

Formula used: When a,b,c are in GP, b² = ac

Proof: When a,b,c,d are in GP then

From the above, we can have the following conclusion

⇒ bc = ad … (i)

⇒ b² = ac … (ii)

⇒ c² = bd … (iii)

Considering (a² – b²), (b² – c²), (c² – d²)

(a² – b²) (c² – d²) = a²c² – a²d² – b²c² + b²d²

= (ac)² – (ad)² – (bc)² + (bd)²

From eqn. (i) , (ii) and (iii)

= (b²)² – (bc)² – (bc)² + (c²)²

= b⁴ – 2b²c² + c⁴

(a² – b²) (c² – d²) = (b² – c²)²

From the above equation we can say that (a² – b²), (b² – c²), (c² – d²) are in GP